# Kinematics Homework: Solving for Velocity and Acceleration

• asi123
In summary, the conversation discusses a problem with finding the angle of a vector using the tangent function. Two possible solutions are suggested, either by drawing a diagram or by using the relationship between tangents and cotangents.

## Homework Statement

In the attachment.

## The Attempt at a Solution

First of all, sorry for the English, I'm from Israel, speak Hebrew here.
Ok, I found v = (-w*R*sin(wt), w*R*cos(wt)), a = (-w^2 * R*cos(wt), -w^2*R*sin(wt)).
I did tg to ay / ax and found the angle of a.
The problem is when I do tg to v, I come out with tg = -ctg, how can I solve that?
Hope u got it, 10x in advance.

#### Attachments

• 1.jpg
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Welcome to PF!

asi123 said:
First of all, sorry for the English, I'm from Israel, speak Hebrew here.

Ok, I found v = (-w*R*sin(wt), w*R*cos(wt)), a = (-w^2 * R*cos(wt), -w^2*R*sin(wt)).
I did tg to ay / ax and found the angle of a.
The problem is when I do tg to v, I come out with tg = -ctg, how can I solve that?

Shalom asi123! Welcome to PF!

(btw, we'd say "I used tg = ay/ax", or "I divided ay by ax to find the tg of the angle of a." )

Two ways to do it …

One … draw a diagram! … that's often a good idea anyway …

in this case, you can see that v is parallel to (-sinωt, cosωt), and a is parallel to (-cosωt, -sinωt), so from ordinary geometry the angle between them is … ?

Two … as you say, the tangent for a is the same as minus the cotangent for v.

Well, ctgθ = tg(what)?

And -tgθ = tg(what)?

(oh … this assumes that ω is constant … are you sure it is?)